Problem: A particle with velocity $v(t)=3\sqrt{t}$, where $t$ is time in seconds, moves in a straight line. How far does the particle move from $t=1$ to $t=4$ seconds?
Solution: The definite integral $ \int_a^b |v(t)| \,dt$ gives the total distance traveled by a particle over the interval $[a,b]$. In this case, $ \int_1^4 |3\sqrt{t}| \,dt$ represents the total distance traveled by the particle from $t=1$ to $t=4$ seconds. Since $3\sqrt{t}$ is always nonnegative, we can rewrite the integral: $\begin{aligned} \int_1^4 |3\sqrt{t}| \,dt&= \int_1^4 3\sqrt{t} \,dt\\ \\ &=3 \int_1^4 \sqrt{t} \,dt\end{aligned}$ We can now evaluate the integral: 3 ∫ 4 1 t √ d t = 3 [ 2 3 x 3 2 ] 4 1 = 3 [ 16 3 − 2 3 ] = 3 ⋅ 14 3 = 14 \begin{aligned}3 \int_1^4 \sqrt{t} \,dt&=3\Big[\dfrac{2}{3}x\^{\frac{3}{2}}\Big]_1^4\\ \\ \\ &=3\Big[\dfrac{16}{3}-\dfrac{2}{3}\Big]\\ \\ &=3\cdot \dfrac{14}{3}\\ \\&=14 \end{aligned} The answer: $14$ units